The proof that (REF ) implies that the metric derivative exists for almost any \(t\)  and belongs to \(L^2(a,b)\)  is in [1]}.
It is easy to prove that any \(\gamma \in H^1(a,b;M)\)  satisfies (REF )
(take \(m(t) = \left| \gamma ^{\prime }(t) \right|_{g}\)  and use (REF )).
We claim that a measurable curve \(\gamma \colon [a, \, b]\rightarrow M\) 
that satisfies (REF ) belongs to \( H^1(a,b;M)\) .
Let \(\gamma \colon [a, \, b]\rightarrow M\)  be a measurable curve
that satisfies (REF ).
We have \(\gamma \in L^2(a, \, b; \, M)\)  because \(M\)  is compact.
Let \(h\in (0, \, b-a)\) .
Equation (REF ) and the Hölder inequality imply
\(\operatorname{dist}_g(\gamma (t), \, \gamma (t+h)) \le h \int _{t}^{t+h} m^2(r) \, \mathrm {d}r\) 
for any \(t\in (a, \, b-h)\) . By Fubini's theorem, we deduce
\(\begin{split}\int _a^{b+h} \operatorname{dist}_g^2(\gamma (t), \, \gamma (t+h)) \, \mathrm {d}t\le h \int _a^{b-h} \left(\int _t^{t+h} m^2(r) \, \mathrm {d}r \right) \mathrm {d}t= h^2 \int _a^{b} m^2(r) \, \mathrm {d}r\end{split}\) 
Therefore, identifying \(M\)  with \(j(M)\subset \mathbb {R}^\ell \) 
we deduce that
\(\int _a^{b+h} \left| \gamma (t) - \gamma (t+h) \right|^2_{\mathbb {R}^\ell } \, \mathrm {d}t\le h^2 \int _a^{b} m^2(r) \, \mathrm {d}r\) 
for any \(h>0\) . This implies \(\gamma \in H^1(a, \, b; \, M)\)  [2]}.
Finally, we show that \(\left| \gamma ^{\prime } \right|_g = \left| \gamma ^{\prime } \right|_g\)  a.e.,
for any \(\gamma \in H^1(a, \, b; \, M)\) .
Let \(\gamma \in H^1(a, \, b; \, M)\) . Let \(s\in (a, \, b)\)  be
a Lebesgue point for the distributional derivative \(\gamma ^\prime \) ,
such that the metric derivative \(\left| \gamma ^{\prime } \right|_g(s)<+\infty \) .
Let \(\sigma >0\)  be smaller than the injectivity radius of \(M\) .
Let \(I(s)\)  be a neighborhood of \(s\)  such that for \(t\in I(s)\) 
there holds \(\gamma (t)\in B_\sigma (\gamma (s))\) . (Such a neighborhood
exists because \(\gamma \)  is continuous, by Sobolev embeddings.)
We consider geodesic normal coordinates centered in \(\gamma (s)\) 
and we let, for \(t\in I(s)\) , \(\bar{\gamma }(t)\in \mathbb {R}^2\) 
be the representation of \(\gamma (t)\)  in this coordinates (see Subsection REF ).
In particular, \(\bar{\gamma }(s) = (0, \, 0)\) .
By definition of geodesic normal coordinates, for any \(t\in I(s)\) 
there exists a unique geodesic of endpoints \(\gamma (s)\)  and \(\gamma (t)\) ,
which is represented in normal coordinates by a straight line segment of
endpoints \(\bar{\gamma }(s) = (0, \, 0)\)  and \(\bar{\gamma }(t)\) .
Moreover, by Gauss' lemma, geodesic normal coordinates
are a radial isometry. Therefore, we have
\(\operatorname{dist}_g(\gamma (t), \, \gamma (s)) = \left| \bar{\gamma }(t) \right|_{\mathbb {R}^2}.\) 
Then,
\(\left| \gamma ^\prime  \right|_g(s)= \lim _{t\rightarrow s}\frac{\operatorname{dist}_g(\gamma (t), \, \gamma (s))}{\left| t-s \right|}= \lim _{t\rightarrow s}\frac{\left| \bar{\gamma }(t) \right|_{\mathbb {R}^2}}{\left| t-s \right|}= \left| \bar{\gamma }^{\prime }(s) \right|_{\mathbb {R}^2}\) 
and, since geodesic normal coordinates centered at \(\gamma (s)\) 
map isometrically \({\gamma (s)} M\)  into \(\mathbb {R}^2\) ,
\(\left| \bar{\gamma }^{\prime }(s) \right|_{\mathbb {R}^2} = \left| \gamma ^{\prime }(s) \right|_g\) .
This completes the proof.
